If vec{a} = 2hat{i} + hat{j} + hat{k},vec{b} | vedclass

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(A) Using the vector triple product formula: $\vec{a} 	imes (\vec{b} 	imes \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec

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$-2, -4, -\frac{2}{3}$

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(A) Using the vector triple product formula: $\vec{a} 	imes (\vec{b} 	imes \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c}$.First,calculate the dot products:$\vec{a} \cdot \vec{c} = (2)(1) + (1)(1) + (1)(2) = 2 + 1 + 2 = 5$.$\vec{a} \cdot \vec{b} = (2)(1) + (1)(2) + (1)(2) = 2 + 2 + 2 = 6$.Now,substitute these into the formula:$\vec{a} 	imes (\vec{b} 	imes \vec{c}) = 5(\hat{i} + 2\hat{j} + 2\hat{k}) - 6(\hat{i} + \hat{j} + 2\hat{k})$$= (5-6)\hat{i} + (10-6)\hat{j} + (10-12)\hat{k} = -\hat{i} + 4\hat{j} - 2\hat{k}$.Comparing this with $(1 + \alpha)\hat{i} + \beta(1 + \alpha)\hat{j} + \gamma(1 + \alpha)(1 + \beta)\hat{k}$:$1 + \alpha = -1 \Rightarrow \alpha = -2$.$\beta(1 + \alpha) = 4 \Rightarrow \beta(1 - 2) = 4 \Rightarrow -\beta = 4 \Rightarrow \beta = -4$.$\gamma(1 + \alpha)(1 + \beta) = -2 \Rightarrow \gamma(1 - 2)(1 - 4) = -2 \Rightarrow \gamma(-1)(-3) = -2 \Rightarrow 3\gamma = -2 \Rightarrow \gamma = -\frac{2}{3}$.Thus,$\alpha = -2, \beta = -4, \gamma = -\frac{2}{3}$.

If $\vec{a} = 2\hat{i} + \hat{j} + \hat{k}$,$\vec{b} = \hat{i} + 2\hat{j} + 2\hat{k}$,$\vec{c} = \hat{i} + \hat{j} + 2\hat{k}$ and $(1 + \alpha)\hat{i} + \beta(1 + \alpha)\hat{j} + \gamma(1 + \alpha)(1 + \beta)\hat{k} = \vec{a} \times (\vec{b} \times \vec{c})$,then $\alpha, \beta, \gamma$ are

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